Question 811796
a. solve {{{2x-3y=4}} for y
{{{2x-3y=4}}}
{{{-3y=-2x+4}}}
{{{y=(2/3)x-(4/3)}}}
The equation can have any y int as long as the slope is {{{2/3}}}
{{{y=(2/3)x+10}}}


b. We already solved this equation for y in the first problem.
y{{{y=(2/3)x-(4/3)}}}
we need a line perpendicular so take the slope of this line and fli it and take its opposite
{{{2/3=-3/2}}}
The equation can have any y int as long as the slope is {{{-3/2}}}
{{{y=(-3/2)x+10}}}


c. We need an equation that passes through both off these points.  find the slope
{{{(5-4)/(3-(-2))}}}
{{{1/5}}}
{{{y=(1/5)x+b}}}
b is the y int.  plug in (3,5) to find b.
{{{5=(1/5)*3+b
{{{5=(3/5)+b}}}
{{{4(2/5)=b}}}
plug back into equation for the answer
{{{y=(1/5)x+4(2/5)}}}


d. find the slope again.
{{{(5-2)/(4-4)}}}
{{{3/0}}}
When there is a zero on the both of the fraction it is undefined.
your equation is
{{{x=0}}}


e.Find the slope.
{{{(-5-(-5))/(7-3)}}}
{{{0/4}}}
When the zero is on the bottom the slope is zero.
Your equation is 
{{{y=o}}}